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3 years ago

What is the answer to this question?

Mathematics

2 answers:

steposvetlana [31]3 years ago

7 0

Answer:

48x - 40y +24

Step-by-step explanation:

Dimas [21]3 years ago

3 0

Answer:

D (48x - 40y +24)

Step-by-step explanation:

First, you need to distribute the 8 throughout the whole equation, so 8×6x=48x, 8×-5y=-40y, and 8×3=24. Right away, you can see that the answer is D.

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3 years ago

(02.01 LC)

ExtremeBDS [4]

Answer:

Yes it is a function

Step-by-step explanation:

We have to check the ordered pairs to find out if given relation is a function or not.

In an ordered pair, the first element represents the input and the second element represents the output.

The set of inputs is domain and output is range.

For a relation to be function, there should be no repetition in domain i.e there should be unique pairs of input and output.

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3 years ago

Is 5 a solution to 2x + 5 = 15? Yes or No<br> How do you know?

dybincka [34]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the measure of the exterior angle?

Nastasia [14]

Answer:

the measure of the exterior angle is 87 degrees.

8 0

2 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago